[POJ2262] Goldbach’s Conjecture
阿新 • • 發佈:2018-06-08
har acc The spa IE () arc ade input Goldbach‘s Conjecture
For example:
Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.)
Anyway, your task is now to verify Goldbach‘s conjecture for all even numbers less than a million.
Each test case consists of one even integer n with 6 <= n < 1000000.
Input will be terminated by a value of 0 for n.
提交地址 : POJ2262
題解 : 怎麽又是水題... 先篩法出素數; 然後暴力判斷;
Code:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 48161 | Accepted: 18300 |
Description
In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:Every even number greater than 4 can bewritten as the sum of two odd prime numbers.
For example:
8 = 3 + 5. Both 3 and 5 are odd prime numbers.
20 = 3 + 17 = 7 + 13.
42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23.
Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.)
Input
The input will contain one or more test cases.Each test case consists of one even integer n with 6 <= n < 1000000.
Input will be terminated by a value of 0 for n.
Output
For each test case, print one line of the form n = a + b, where a and b are odd primes. Numbers and operators should be separated by exactly one blank like in the sample output below. If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized. If there is no such pair, print a line saying "Goldbach‘s conjecture is wrong."Sample Input
8 20 42 0
Sample Output
8 = 3 + 5 20 = 3 + 17 42 = 5 + 37
Source
Ulm Local 1998提交地址 : POJ2262
題解 : 怎麽又是水題... 先篩法出素數; 然後暴力判斷;
Code:
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 5 bool isprime[1000010]; //1 -> 合數, 0 -> 質數 6 7 int main() 8 { 9 isprime[0] = isprime[1] = 1; 10 for (register int i = 2 ; i <= 1000001 ; i ++) 11 { 12 if (isprime[i]) continue; 13 for (register int j = i ; j <= 1000001/i ; j ++) 14 isprime[i*j] = 1; 15 } 16 17 int n; 18 while (scanf("%d", &n) != EOF) 19 { 20 if (n == 0) return 0; 21 for (register int i = 0 ; i <= n ; i ++) 22 { 23 if (!isprime[i] and !isprime[n-i]) 24 { 25 printf("%d = %d + %d\n", n, i, n-i); 26 break; 27 } 28 } 29 } 30 31 }
[POJ2262] Goldbach’s Conjecture